I'm trying to prove the following by induction but I'm stuck.
$x_1 = 1, x_2 = 2, x_n=\frac{1}{2}(x_{n-1}+x_{n-2})$.
Show that: $x_n-x_{n+1} = \frac{(-1)^n}{2^{n-1}}$
I proved the basis step, but I'm stuck in the inductive step. I tried going from L.H.S and take as $x_{n+2}$ as common divisor and had $(x_{n-1} - 1)$. I didn't know where to go from there..
Hint:
Note you have, using the recurrence relation,
$$\begin{equation}\begin{aligned} x_{n+1} - x_{n+2} & = x_{n+1} - \frac{1}{2}(x_{n+1} + x_{n}) \\ & = x_{n+1} - \frac{x_{n+1}}{2} - \frac{x_{n}}{2} \\ & = \frac{1}{2}(x_{n+1} - x_{n}) \\ & = -\frac{1}{2}(x_{n} - x_{n+1}) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$